Chapter B, Problem B.74P

To determine

(a)

The principal mass moment of inertias at the origin.

Answer to Problem B.74P

The principal mass moment of inertias at the origin are $0.22583{\text{kg-m}}^{\text{2}}$, $0.41920{\text{kg-m}}^{\text{2}}$ and $0.51621{\text{kg-m}}^{\text{2}}$.

Explanation of Solution

Given information:

The mass per unit length of the steel is $0.056\text{kg}/\text{m}$.

Draw the diagram for the different section of the body.

Figure-(1)

Concept used:

Write the expression for the mass of each section.

$m=\left(m_L\right)L$   ....... (I)

Here, the mass per unit length is $m_L$ and the length of the side is $L$.

Write the expression of mass moment of inertia of section 1 about $x$ -axis.

${\left(I_x\right)}_1=m\left[\frac{1}{3}{L}^2\right]$   ....... (II)

Write the expression of mass moment of inertia of section 2 about $x$ -axis.

${\left(I_x\right)}_2=m\left(L^2\right)$   ....... (III)

Write the expression of total mass moment of inertia about $x$ -axis using parallel axis theorem.

$I_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_3+{\left(I_x\right)}_4+{\left(I_x\right)}_5+{\left(I_x\right)}_6$   ....... (IV)

Here, the mass moment of inertia for section 3 about x- axis is ${\left(I_x\right)}_3$, the mass moment of inertia for section 4 about x- axis is ${\left(I_x\right)}_4$, the mass moment of inertia for section 5about x- axis is ${\left(I_x\right)}_5$ and the mass moment of inertia for section 6about x- axis is ${\left(I_x\right)}_6$, the total mass moment of inertia of all the components about $x$ - axis is $I_x$.

The mass moment of inertia for section 1 is equal to the mass moment of inertia for section 3.

The mass moment of inertia for section 1 is equal to the mass moment of inertia for section 4.

The mass moment of inertia for section 1 is equal to the mass moment of inertia for section 6.

The mass moment of inertia for section 2 is equal to the mass moment of inertia for section 5.

Substitute ${\left(I_x\right)}_2$ for ${\left(I_x\right)}_5$, ${\left(I_x\right)}_1$ for ${\left(I_x\right)}_3$, ${\left(I_x\right)}_1$ for ${\left(I_x\right)}_4$ and ${\left(I_x\right)}_1$ for ${\left(I_x\right)}_6$ in Equation (IV).

$\begin{array}{c}{I}_x={\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_1+{\left(I_x\right)}_1+{\left(I_x\right)}_2+{\left(I_x\right)}_1\\ =4{\left(I_x\right)}_1+2{\left(I_x\right)}_2\end{array}$   ....... (V)

Write the expression of total mass moment of inertia about $y$ -axis using parallel axis theorem.

$I_y={\left(I_y\right)}_1+{\left(I_y\right)}_2+{\left(I_y\right)}_3+{\left(I_y\right)}_4+{\left(I_y\right)}_5+{\left(I_y\right)}_6$   ....... (VI)

Here, the mass moment of inertia for section 3 about y- axis is ${\left(I_y\right)}_3$, the mass moment of inertia for section 4 about y- axis is ${\left(I_y\right)}_4$, the mass moment of inertia for section 5about y- axis is ${\left(I_y\right)}_5$ and the mass moment of inertia for section 6about y- axis is ${\left(I_y\right)}_6$, the total mass moment of inertia of all the components about $y$ - axis is $I_y$.

The mass moment of inertia for section 1 is equal to zero.

The mass moment of inertia for section 4 is equal to the mass moment of inertia for section 5.

The mass moment of inertia for section 2 is equal to the mass moment of inertia for section 6.

Substitute ${\left(I_y\right)}_4$ for ${\left(I_y\right)}_5$, ${\left(I_y\right)}_2$ for ${\left(I_y\right)}_6$ and $0$ for ${\left(I_y\right)}_1$ in Equation (IV).

$\begin{array}{c}{I}_y=0+{\left(I_y\right)}_2+{\left(I_y\right)}_3+{\left(I_y\right)}_4+{\left(I_y\right)}_4+{\left(I_y\right)}_2\\ =2{\left(I_y\right)}_2+{\left(I_y\right)}_3+2{\left(I_y\right)}_4\end{array}$   ....... (VII)

Write the expression of mass moment of inertia of section 2 about $y$ -axis.

${\left(I_y\right)}_2=m\left[\frac{1}{3}{L}^2\right]$   ....... (VIII)

Write the expression of mass moment of inertia of section 3 about $y$ -axis.

${\left(I_y\right)}_3=m\left(L^2\right)$   ....... (IX)

Write the expression of mass moment of inertia of section 4 about $y$ -axis.

${\left(I_y\right)}_4=m\left(\left(\frac{1}{12}{L}^2+L^2+{\left(\frac{L}{2}\right)}^2\right)\right)$   ....... (X)

The figure below illustrates the centroidal axis of a component.

Figure-(2)

From the symmetry in above figure about $x^{\prime }$ - $y^{\prime }$ plane.

${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1={\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_2={\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_3={\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_4={\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_5={\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_6=0$   ....... (XI)

Here, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 1 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1$, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 2 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_2$, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 3 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_3$, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 4 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_4$, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 5 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_5$, and the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane for component 6 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_6$

From the symmetry in the above figure about $y^{\prime }$ - $z^{\prime }$ plane.

${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_2={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_3={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_4={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_5={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_6=0$   ....... (XII)

Here, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 1 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 2 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_2$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 3 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_3$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 4 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_4$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 5 is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_5$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane for component 6 plane is ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_6$.

From the symmetry in the above figure about $z^{\prime }$ - $x^{\prime }$ plane.

${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_1={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_2={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_3={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_4={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_5={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_6=0$   ....... (XII)

Here, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 1 is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_1$, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 2 is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_2$, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 3 is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_3$, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 4 is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_4$, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 5 is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_5$, and the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane for component 6 plane is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_6$.

Write the expression for product of mass moment of inertia in $x$ - $y$ plane.

$\begin{array}{c}\left(I_{\text{xy}}\right)=\sum _{\text{i}=1}^{\text{i}=5}\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_{\text{i}}+m\overline{x_{\text{i}}}\overline{y_{\text{i}}}\right)\\ \left(I_{\text{xy}}\right)=\left[\begin{array}{l}\left({\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1+m\overline{x_1}\overline{y_1}\right)+\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_2+m\overline{x_2}\overline{y_2}\right)\\ +\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_3+m\overline{x_3}\overline{y_3}\right)+\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_4+m\overline{x_4}\overline{y_4}\right)\\ +\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_5+m\overline{x_5}\overline{y_5}\right)+\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_6+m\overline{x_6}\overline{y_6}\right)\end{array}\right]\end{array}$   ....... (XIII)

Here, the product mass moment of inertia is $\left(I_{\text{xy}}\right)$, the subscript limit is $\text{i}$, the distance between the centroid and $x$ -axis of component 1 is $\overline{x_1}$, the distance between the centroid and $y$ -axis of component 1 is $\overline{y_1}$, the distance between the centroid and $x$ -axis of component 2 is $\overline{x_2}$, the distance between the centroid and $y$ -axis of component 2 is $\overline{y_2}$, the distance between the centroid and $x$ -axis of component 3 is $\overline{x_3}$, the distance between the centroid and $y$ -axis of component 3 is $\overline{y_3}$, the distance between the centroid and $x$ -axis of component 4 is $\overline{x_4}$, the distance between the centroid and $y$ -axis of component 4 is $\overline{y_4}$, the distance between the centroid and $x$ -axis of component 5 is $\overline{x_5}$, the distance between the centroid and $y$ -axis of component 5 is $\overline{y_5}$, the distance between the centroid and $x$ -axis of component 6 is $\overline{x_6}$, the distance between the centroid and $y$ -axis of component 6 is $\overline{y_6}$.

Write the expression for product mass moment of inertia in $y$ - $z$ plane.

$\begin{array}{c}\left(I_{\text{yz}}\right)=\sum _{\text{i}=1}^{\text{i}=5}\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><mo> z'</mo>}}\right)}_{\text{i}}+m_{\text{i}}\overline{y_{\text{i}}}\overline{z_{\text{i}}}\right)\\ \left(I_{\text{yz}}\right)=\left[\begin{array}{l}\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1+m\overline{y_1}\overline{z_1}\right)+\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_2+m\overline{y_2}\overline{z_2}\right)\\ +\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_3+m\overline{y_3}\overline{z_3}\right)+\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_4+m\overline{y_4}\overline{z_4}\right)\\ +\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_5+m\overline{y_5}\overline{z_5}\right)+\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_6+m\overline{y_6}\overline{z_6}\right)\end{array}\right]\end{array}$   ....... (XIV)

Here, the product mass moment of inertia is $\left(I_{\text{yz}}\right)$, the subscript limit is $\text{i}$, the distance between the centroid and $y$ -axis of component 1 is $\overline{y_1}$, the distance between the centroid and $z$ -axis of component 1 is $\overline{z_1}$, the distance between the centroid and $y$ -axis of component 2 is $\overline{y_2}$, the distance between the centroid and $z$ -axis of component 2 is $\overline{z_2}$, the distance between the centroid and $y$ -axis of component 3 is $\overline{y_3}$, the distance between the centroid and $z$ -axis of component 3 is $\overline{z_3}$, the distance between the centroid and $y$ -axis of component 4 is $\overline{y_4}$, the distance between the centroid and $z$ -axis of component 4 is $\overline{z_4}$, the distance between the centroid and $y$ -axis of component 5 is $\overline{y_5}$, the distance between the centroid and $z$ -axis of component 5 is $\overline{z_5}$, the distance between the centroid and $y$ -axis of component 6 is $\overline{y_6}$, the distance between the centroid and $z$ -axis of component 6 is $\overline{z_6}$.

Write the expression for product mass moment of inertia in $z$ - $x$ plane.

$\begin{array}{c}\left(I_{\text{zx}}\right)=\sum _{\text{i}=1}^{\text{i}=5}\left({\left(I_{\text{z'<msup><mo> x</mo><mo>′</mo></msup>}}\right)}_{\text{i}}+m\overline{x_{\text{i}}}\overline{z_{\text{i}}}\right)\\ \left(I_{\text{zx}}\right)=\left[\begin{array}{l}\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_1+m\overline{x_1}\overline{z_1}\right)+\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_2+m\overline{x_2}\overline{z_2}\right)\\ +\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_3+m\overline{x_3}\overline{z_3}\right)+\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_4+m\overline{x_4}\overline{z_4}\right)\\ +\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_5+m\overline{x_5}\overline{z_5}\right)+\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_6+m\overline{x_6}\overline{z_6}\right)\end{array}\right]\end{array}$   ....... (XV)

Here, the product mass moment of inertia in $z$ - $x$ plane is $\left(I_{\text{zx}}\right)$, the distance between the centroid and $x$ -axis of component 1 is $\overline{x_1}$, the distance between the centroid and $z$ -axis of component 1 is $\overline{z_1}$, the distance between the centroid and $x$ -axis of component 2is $\overline{x_2}$, the distance between the centroid and $z$ -axis of component 2 is $\overline{z_2}$, the distance between the centroid and $x$ -axis isof component 3is $\overline{x_3}$, the distance between the centroid and $z$ -axis isof component 3is $\overline{z_3}$, the distance between the centroid and $x$ -axis of component 4is $\overline{x_4}$, the distance between the centroid and $z$ -axis of component 4is $\overline{z_4}$, the distance between the centroid and $x$ -axis isof component 5is $\overline{x_5}$, the distance between the centroid and $z$ -axis isof component 5 is $\overline{z_5}$, the distance between the centroid and $x$ -axis is of component 6 is $\overline{x_6}$, the distance between the centroid and $z$ -axis is of component 6 is $\overline{z_6}$.

Write the expression of mass moment of inertia with respect o origin along the unit vector $\lambda$.

$I=I_{\text{x}}{\lambda }_{\text{x}}^2+I_{\text{y}}{\lambda }_{\text{y}}^2+I_{\text{z}}{\lambda }_{\text{z}}^2-2I_{\text{xy}}{\lambda }_{\text{x}}{\lambda }_{\text{y}}-2I_{\text{yz}}{\lambda }_z{\lambda }_{\text{y}}-2I_{\text{xz}}{\lambda }_{\text{x}}{\lambda }_z$   ....... (XVI)

Here, the mass moment of inertia with respect to origin along unit vector $\lambda$ is $I$, the coefficient of $\hat{i}$ in unit vector $\lambda$ is ${\lambda }_{\text{x}}$, the coefficient of $\hat{j}$ in unit vector $\lambda$ is ${\lambda }_{\text{y}}$, the coefficient of $\hat{k}$ in unit vector $\lambda$ is ${\lambda }_z$.

Calculation:

Substitute $0.056\text{kg}/\text{m}$ for $m_L$ and $1.2\text{m}$ for $L$ in Equation (I).

$\begin{array}{c}m=\left(0.056\text{kg}/\text{m}\right)\left(1.2\text{m}\right)\\ =0.0672\text{kg}\end{array}$

Substitute $0.0672\text{kg}$ for $m$ and $1.2\text{m}$ for $L$ in Equation (II).

$\begin{array}{c}{\left(I_x\right)}_1=0.0672\text{kg}\left[\frac{1}{3}{\left(1.2\text{m}\right)}^2\right]\\ =0.0672\text{kg}\left[\frac{1}{3}\left(1.44{\text{m}}^2\right)\right]\\ =0.03225\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.0672\text{kg}$ for $m$ and $1.2\text{m}$ for $L$ in Equation (III).

$\begin{array}{c}{\left(I_x\right)}_2=0.0672\text{kg}{\left(1.2\text{m}\right)}^2\\ =0.0672\text{kg}\left(1.44{\text{m}}^2\right)\\ =0.09676\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.09676\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_2$ and $0.03225\text{kg}\cdot {\text{m}}^2$ for ${\left(I_x\right)}_1$ in Equation (V).

$\begin{array}{c}{I}_x=4\left(0.03225\text{kg}\cdot {\text{m}}^2\right)+2\left(0.09676\text{kg}\cdot {\text{m}}^2\right)\\ =\left(0.129\text{kg}\cdot {\text{m}}^2\right)+\left(0.1935\text{kg}\cdot {\text{m}}^2\right)\\ =0.322\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.0672\text{kg}$ for $m$ and $1.2\text{m}$ for $L$ in Equation (VIII).

$\begin{array}{c}{\left(I_y\right)}_2=0.0672\text{kg}\left[\frac{1}{3}{\left(1.2\text{m}\right)}^2\right]\\ =0.0672\text{kg}\left[\frac{1}{3}\left(1.44{\text{m}}^2\right)\right]\\ =0.03225\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.0672\text{kg}$ for $m$ and $1.2\text{m}$ for $L$ in Equation (IX).

$\begin{array}{c}{\left(I_y\right)}_3=0.0672\text{kg}{\left(1.2\text{m}\right)}^2\\ =0.0672\text{kg}\left(1.44{\text{m}}^2\right)\\ =0.09676\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.0672\text{kg}$ for $m$ and $1.2\text{m}$ for $L$ in Equation (X).

$\begin{array}{c}{\left(I_y\right)}_4=0.0672\text{kg}\left(\left(\frac{1}{12}{\left(1.2\text{m}\right)}^2+{\left(1.2\text{m}\right)}^2+{\left(\frac{1.2\text{m}}{2}\right)}^2\right)\right)\\ =0.0672\text{kg}\left(\left(\left(\frac{1}{12}1.44{\text{m}}^2\right)+\left(1.44{\text{m}}^2\right)+\left(0.36{\text{m}}^2\right)\right)\right)\\ =0.0672\text{kg}\left(1.92{\text{m}}^2\right)\\ =0.1290\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.1290\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_4$, $0.09676\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_3$ and $0.03225\text{kg}\cdot {\text{m}}^2$ for ${\left(I_y\right)}_2$ in Equation (VII).

$\begin{array}{c}{I}_y=2\left(0.03225\text{kg}\cdot {\text{m}}^2\right)+\left(0.09676\text{kg}\cdot {\text{m}}^2\right)+2\left(0.1290\text{kg}\cdot {\text{m}}^2\right)\\ =\left(0.06450\text{kg}\cdot {\text{m}}^2\right)+\left(0.09676\text{kg}\cdot {\text{m}}^2\right)+\left(0.258\text{kg}\cdot {\text{m}}^2\right)\\ =0.4193\text{kg}\cdot {\text{m}}^2\end{array}$

The mass moment of inertia about z- axis and the mass moment of inertia about y- axis is equal due to symmetry.

Hence, $I_z=0.4193\text{kg}\cdot {\text{m}}^2$.

Substitute $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_2$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_3$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_4$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_5$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_6$, $0.0672\text{kg}$ for $m$, $0.6\text{m}$ for $\overline{y_1}$, $0\text{m}$ for $\overline{x_1}$, $0.6\text{m}$ for $\overline{x_2}$, $1.2\text{m}$ for $\overline{y_2}$, $1.2\text{m}$ for $\overline{x_3}$, $0.6\text{m}$ for $\overline{y_3}$, $1.2\text{m}$ for $\overline{x_4}$, $0\text{m}$ for $\overline{y_4}$, $0.6\text{m}$ for $\overline{x_5}$, $0\text{m}$ for $\overline{y_5}$, $0\text{m}$ for $\overline{x_6}$, $0\text{m}$ for $\overline{y_6}$ in Equation (XIII).

$\begin{array}{c}\left(I_{\text{xy}}\right)=\left[\begin{array}{l}\left(0+\left(0.0672\text{kg}\right)0\left(0.6\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(0.6\text{m}\right)\left(1.2\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(0.6\text{m}\right)\left(0\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(0\text{m}\right)\left(0\text{m}\right)\right)\end{array}\right]\\ =2\left(\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\right)\\ =0.096768\text{kg}/{\text{m}}^2\end{array}$

Substitute $0$ for ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1$, $0$ for ${\left(I_{{}_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}}\right)}_2$, $0$ for ${\left(I_{{}_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}}\right)}_3$, $0$ for ${\left(I_{{}_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}}\right)}_4$, $0$ for ${\left(I_{{}_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}}\right)}_5$, $0$ for ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_6$, $0.0672\text{kg}$ for $m$, $0.6\text{m}$ for $\overline{y_1}$, $0\text{m}$ for $\overline{z_1}$, $0\text{m}$ for $\overline{z_2}$, $1.2\text{m}$ for $\overline{y_2}$, $0\text{m}$ for $\overline{z_3}$, $0.6\text{m}$ for $\overline{y_3}$, $0.6\text{m}$ for $\overline{z_4}$, $0\text{m}$ for $\overline{y_4}$, $1.2\text{m}$ for $\overline{z_5}$, $0\text{m}$ for $\overline{y_5}$, $0.6\text{m}$ for $\overline{x_6}$, $0\text{m}$ for $\overline{y_6}$ in Equation (XIV).

$\begin{array}{c}\left(I_{\text{yz}}\right)=\left[\begin{array}{l}\left(0+\left(0.0672\text{kg}\right)0\left(0.6\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(0.6\text{m}\right)\left(0\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\left(0\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(0\text{m}\right)\left(0.6\text{m}\right)\right)\end{array}\right]\\ =0\text{kg}/{\text{m}}^2\end{array}$

Substitute $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_1$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_2$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_3$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_4$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_5$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup>}{x}^{\prime }}\right)}_6$, $0.0672\text{kg}$ for $0\text{m}$ for $\overline{z_1}$, $0\text{m}$ for $\overline{x_1}$, $0.6\text{m}$ for $\overline{x_2}$, $0\text{m}$ for $\overline{z_2}$, $1.2\text{m}$ for $\overline{x_3}$, $0\text{m}$ for $\overline{z_3}$, $1.2\text{m}$ for $\overline{x_4}$, $0.6\text{m}$ for $\overline{z_4}$, $0.6\text{m}$ for $\overline{x_5}$, $1.2\text{m}$ for $\overline{z_5}$, $0\text{m}$ for $\overline{x_6}$, $0.6\text{m}$ for $\overline{z_6}$ in Equation (XV).

$\begin{array}{c}\left(I_{\text{zx}}\right)=\left[\begin{array}{l}\left(0+\left(0.0672\text{kg}\right)\left(0\text{m}\right)\left(0\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(0.6\text{m}\right)\left(0\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\right)\\ +\left(0+\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\right)+\left(0+\left(0.0672\text{kg}\right)\left(0\text{m}\right)\left(0.6\text{m}\right)\right)\end{array}\right]\\ =2\left(\left(0.0672\text{kg}\right)\left(1.2\text{m}\right)\left(0.6\text{m}\right)\right)\\ =0.096768\text{kg}/{\text{m}}^2\end{array}$ The principal mass moment of inertia at the origin is calculated as follows:

$\begin{array}{l}{K}^3-\left(I_x+I_y+I_z\right)K^2+\left(2I_x{I}_y+{\left(I_y\right)}^2-2{\left(I_{xy}\right)}^2\right)K\\ -\left(I_x{I}_y{I}_z-2I_y{I}_{zx}\right)=0\\ K^3-\left(0.32258+2\times 0.41933\right)K^2+\\ \left(2\left(0.32258\right)\left(0.41933\right)+{\left(0.41933\right)}^2-2{\left(0.096768\right)}^2\right)K\\ -\left(\left(0.32258\right){\left(0.41933\right)}^2-2\left(0.41933\right)\left(0.096768\right)\right)=0\\ K^3-\left(1.16124\right)K^2+\left(0.42764\right)K-0.048869=0\end{array}$

After solving the above equation,

$\begin{array}{l}{K}_1=0.22583{\text{kg-m}}^{\text{2}}\\ K_2=0.41920{\text{kg-m}}^{\text{2}}\\ \text{and}\\ K_3=0.51621{\text{kg-m}}^{\text{2}}\end{array}$

Conclusion:

The principal mass moment of inertias at the origin are $0.22583{\text{kg-m}}^{\text{2}}$, $0.41920{\text{kg-m}}^{\text{2}}$ and $0.51621{\text{kg-m}}^{\text{2}}$.

To determine

(b)

The principal axis about the origin.

Answer to Problem B.74P

The principal axis about the origin are ${\left({\theta }_x\right)}_1=35.3°$, ${\left({\theta }_y\right)}_1=65.9°$, ${\left({\theta }_z\right)}_1=65.9°$, ${\left({\theta }_x\right)}_2=90°$, ${\left({\theta }_y\right)}_2=45°$, ${\left({\theta }_z\right)}_2=135°$, ${\left({\theta }_x\right)}_3=54.7°$, ${\left({\theta }_y\right)}_3=125.3°$ and ${\left({\theta }_z\right)}_3=125.3°$.

Explanation of Solution

Given information:

The mass per unit length of the steel is $0.056\text{kg}/\text{m}$.

Draw the diagram for the different section of the body.

Figure-(1)

Calculation:

The direction cosine is calculated as follows:

$\begin{array}{l}-I_{xy}{\left({\lambda }_x\right)}_1+\left(I_y-K_1\right){\left({\lambda }_y\right)}_1-I_{yz}{\left({\lambda }_z\right)}_3=0\\ \text{and}\\ -I_{zx}{\left({\lambda }_x\right)}_1-I_{yz}{\left({\lambda }_z\right)}_1+\left(I_y-K_1\right){\left({\lambda }_y\right)}_3=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(-0.096768\right){\left({\lambda }_x\right)}_1+\left(0.41933-0.22583\right){\left({\lambda }_y\right)}_1=0\\ \text{and}\\ \left(-0.096768\right){\left({\lambda }_x\right)}_1+\left(0.41933-0.22583\right){\left({\lambda }_z\right)}_1=0\end{array}$

After solving above equations,

${\left({\lambda }_y\right)}_1={\left({\lambda }_z\right)}_1=0.50009{\left({\lambda }_x\right)}_1$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}2{\left(0.50009{\left({\lambda }_x\right)}_1\right)}^2+{\left({\lambda }_x\right)}^2{}_1=1\\ {\left({\lambda }_x\right)}_1=0.81645\end{array}$

Direction cosine in y and z direction are,

${\left({\lambda }_y\right)}_1={\left({\lambda }_z\right)}_1=0.40830$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_1=35.3°\\ {\left({\theta }_y\right)}_1={\left({\theta }_z\right)}_1=65.9°\end{array}$

Again,

The direction cosine is calculated as follows:

$\begin{array}{l}\left(I_x-K_2\right){\left({\lambda }_x\right)}_2+\left(I_{xy}\right){\left({\lambda }_y\right)}_2-I_{zx}{\left({\lambda }_z\right)}_2=0\\ \text{and}\\ -I_{xy}{\left({\lambda }_x\right)}_2+\left(I_y-K_2\right){\left({\lambda }_y\right)}_2-\left(I_{yz}\right){\left({\lambda }_z\right)}_2=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(0.32258-0.41920\right){\left({\lambda }_x\right)}_2-\left(0.096768\right){\left({\lambda }_y\right)}_2-\left(0.096768\right){\left({\lambda }_z\right)}_2=0\\ \text{and}\\ \left(-0.96768\right){\left({\lambda }_x\right)}_2+\left(0.41933-0.41920\right){\left({\lambda }_y\right)}_2=0\end{array}$

After solving above equations,

$\begin{array}{l}{\left({\lambda }_x\right)}_2=0\\ \text{and}\\ {\left({\lambda }_z\right)}_2=-{\left({\lambda }_y\right)}_2\end{array}$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}{\left(0\right)}^2+{\left({\lambda }_y\right)}^2{}_2+{\left(-{\left({\lambda }_y\right)}_2\right)}^2=1\\ {\left({\lambda }_y\right)}_2=\frac{1}{\sqrt{2}}\end{array}$

Direction cosine in z direction is,

${\left({\lambda }_z\right)}_2=\frac{-1}{\sqrt{2}}$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_2=90°\\ {\left({\theta }_y\right)}_2=45°\\ \text{and}\\ {\left({\theta }_z\right)}_2=135°\end{array}$

Similarly,

The direction cosine is calculated as follows:

$\begin{array}{l}\left(I_{xy}\right){\left({\lambda }_x\right)}_3+\left(I_y-K_3\right){\left({\lambda }_y\right)}_3+I_{yz}{\left({\lambda }_z\right)}_3=0\\ \text{and}\\ -I_{zx}{\left({\lambda }_x\right)}_3-I_{yz}{\left({\lambda }_y\right)}_3+\left(I_z-K_3\right){\left({\lambda }_z\right)}_3=0\end{array}$

Substitute the values from the sub-part (a) in above equations as follows:

$\begin{array}{l}\left(-0.096768\right){\left({\lambda }_x\right)}_3+\left(0.41933-0.51621\right){\left({\lambda }_y\right)}_3=0\\ \text{and}\\ \left(-0.096768\right){\left({\lambda }_x\right)}_3+\left(0.41933-0.51621\right){\left({\lambda }_z\right)}_3=0\end{array}$

After solving above equations,

${\left({\lambda }_y\right)}_3={\left({\lambda }_z\right)}_3=-{\left({\lambda }_x\right)}_3$

Direction cosine in x direction is calculated as follows:

$\begin{array}{c}{\left({\left({\lambda }_x\right)}_3\right)}^2+2{\left(-{\left({\lambda }_x\right)}_3\right)}^2=1\\ {\left({\lambda }_x\right)}_3=\frac{1}{\sqrt{3}}\end{array}$

Direction cosine in y and z direction are,

${\left({\lambda }_y\right)}_3={\left({\lambda }_z\right)}_3=-\frac{1}{\sqrt{3}}$

So, the direction is calculated as follows:

$\begin{array}{l}{\left({\theta }_x\right)}_3=54.7°\\ {\left({\theta }_y\right)}_3={\left({\theta }_z\right)}_3=125.3°\end{array}$

The sketch is shown below:

Conclusion:

So, the principal axis about the origin are ${\left({\theta }_x\right)}_1=35.3°$, ${\left({\theta }_y\right)}_1=65.9°$, ${\left({\theta }_z\right)}_1=65.9°$, ${\left({\theta }_x\right)}_2=90°$, ${\left({\theta }_y\right)}_2=45°$, ${\left({\theta }_z\right)}_2=135°$, ${\left({\theta }_x\right)}_3=54.7°$, ${\left({\theta }_y\right)}_3=125.3°$ and ${\left({\theta }_z\right)}_3=125.3°$.